1965 AMC 12 第 29 题

先试着解答 1965 AMC 12 第 29 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1965 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

29.

在至少选修一门课程的 2828 名学生中,只选数学和英语的人数等于只选数学的人数。没有学生只选英语或只选历史,有六名学生选数学和历史但不选英语。只选英语和历史的人数是三门都选人数的五倍。若三门都选的人数为非零偶数,则只选英语和数学的人数为:
Of 2828 students taking at least one subject the number taking Mathematics and English only equals the number taking Mathematics only. No student takes English only or History only, and six students take Mathematics and History, but no English. The number taking English and History only is five times the number taking all three subjects. If the number taking all three subjects is even and non-zero, the number taking English and Mathematics only is:
55
66
77
88
99
答案:A
知识点:韦恩图基本计数丢番图方程
小提示:
设三门都选的人数为 tt,只选数学和只选数学英语的人数均为 xx
Let tt be the all-three count and xx both the Math-only and Math-English-only count
大提示:
总人数可写成 2x+6+6t=282x+6+6t=28
The total becomes 2x+6+6t=282x+6+6t=28
解答:
设三门都选的人数为 tt,只选数学和英语的人数为 xx,它也等于只选数学的人数。各不相交区域的人数之和为 x+x+6+5t+t=28,x+x+6+5t+t=28\text{,} 所以 x=11−3tx=11-3t。 因为 tt 为正偶数,所以只有 t=2t=2 能使所列人数非负,此时 x=5x=5。
因此,正确答案是 A。
Let tt be the number taking all three and xx the number taking Mathematics and English only, which also equals the Mathematics-only count. The disjoint-region total is x+x+6+5t+t=28,x+x+6+5t+t=28, so x=11−3t.x=11-3t. Since tt is positive and even, t=2t=2 is the only value giving a nonnegative listed count, and x=5.x=5.
Therefore, the correct answer is A.
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